I need help on this X-linked Biology question

Question

A human female "carrier" who is heterozygous for the recessive, sex-linked trait causing red-green color blindness, marries a normal male. What proportion of their male progeny will have red-green color blindness? Show a Punnett square 4. In my last microbiology course a female student said she was X-linked color blind (a recessive trait) and was having trouble, determining the results of an experiment. I asked this student if she had any sons, and she had three. What are the phenotypes of he sons, regardless of the father? First figure out her genotype. Show a Punnett square to prove your result

Answer 1

Let me know if this helps!

First question:

1. say that X^R indicates normal vision (dominant) and X^r indicates colorblind vision (recessive). The mother is X^RX^r because she is a heterozygous carrier
2. The father, a normal male, is X^RY
3. Therefore the Punnet Square is X^RX^r (mother) crossed with X^RY (father)
4. See if you are able to figure out the rest of the problem with this information!

Second Question:

1. The mother is a female with colorblindness. Because colorblindness is recessive, her genotype must be X^rX^r
2. It does not matter what the fathers genotype is, because any male progeny will have an XY genotype where the Y allele is from the father and the X allele is from the mother. Because the X allele in the male will always be from the mother, all sons will have a genotype of X^rY, meaning that all sons will have a colorblind phenotype

I tried to upload a photo of a Punnet square, but was unable to. Please contact me if you need more help with a Punnet square drawing!